$\sum p(|S_n|\geq n\epsilon)<\infty\Leftrightarrow EX_1=0,E[X_1^2]<\infty$ $\{X_n\}$ are independent identical random variables, $S_n=X_1+...+X_n$.
Proof: for each $\epsilon>0,$
\begin{equation} 
\sum p(|S_n|\geq n\epsilon)<\infty\Leftrightarrow EX_1=0,E[X_1^2]<\infty
\end{equation}
My ideas so far:
From Borel 0-1 law:
\begin{equation} 
\sum p(|S_n|\geq n\epsilon)<\infty\Leftrightarrow p(\text{limsup}\frac{|Sn|}{n}\geq \epsilon )=0\Leftrightarrow \text{lim} \frac{|S_n|}{n}=0\Leftrightarrow\text{lim} \frac{S_n}{n}=0
\end{equation}
Also, $\sum p(|S_n|\geq n\epsilon)<\infty\Leftrightarrow E|S_n|<\infty$ (From the formula $\sum p(|S_n|\geq n)\leq E|S_n|\leq 1+\sum p(|S_n|\geq n)$), then $E|X_n|<\infty$. By strong law of large number, we have:
\begin{equation}
0=\text{lim}\frac{Sn}{n}=EX_1
\end{equation}
Which means:
$\sum p(|S_n|\geq n\epsilon)<\infty\Leftrightarrow EX_1=0$.
I don't known how to prove or use the condition $E[X_1^2]<\infty$. Are there something wrong with my proof?
Thanks in advance for any tips or help in general.
 A: Suppose that $\sum \mathbb P(|S_n|\geq n\epsilon)<\infty$ for each positive $\varepsilon$. Then $S_n/n\to 0$ almost surely, hence $X_n/n\to 0$ almost surely and it follows that $\mathbb E\left\lvert X_1\right\rvert<\infty$ and by the law of large numbers, $\mathbb E\left[X_1\right]=0$.
We use Ottaviani's inequality: if we put $M_k:=\max_{1\leqslant i\leqslant k}|S_i|$ and $S_{k,n}:=\sum_{i=k}^nX_i$, then for all $\varepsilon >0$ we have
$$\min_{1\leqslant k\leqslant n}\mathbb P(|S_{k,n}|\leqslant\varepsilon)\mathbb P(|M_n|>2\varepsilon)\leqslant \mathbb P(|S_n|>\varepsilon).$$
Using this with $\varepsilon$ replaced by $n\varepsilon$ and noticing that there exists $c>0$ and an integer $n_0$ such that
$\min_{1\leqslant k\leqslant n}\mathbb P(|S_{k,n}|\leqslant\varepsilon)>c$ for $n\geqslant n_0$, we derive that
$$\sum \mathbb P\left(\max_{1\leqslant k\leqslant n}|S_k|> n\epsilon\right)<\infty,$$
which implies that for each positive $\varepsilon$,
$$\sum \mathbb P\left(\max_{1\leqslant k\leqslant n}|X_k|> n\epsilon\right)<\infty.$$
Using Bonferroni's inequality, independence and the fact that $n\mathbb P\left(\left\lvert X_1\right\rvert>n\varepsilon\right)\to 0$, we derive that
$$\sum n\mathbb P\left(|X_1|> n\epsilon\right)<\infty,$$
giving square integrability.
For the converse, use truncation at level $n$, that is, view $S_n$ as
$S_n'+S''_n$, with
$$
S_n'=\sum_{k=1}^nX_k\mathbf{1}_{\{\lvert X_k\rvert \leqslant n\}}-\mathbb E\left[X_k\mathbf{1}_{\{\lvert X_k\rvert \leqslant n\}}\right]
$$
$$
S_n''=\sum_{k=1}^nX_k\mathbf{1}_{\{\lvert X_k\rvert \gt n\}}-\mathbb E\left[X_k\mathbf{1}_{\{\lvert X_k\rvert \gt n\}}\right].
$$
